• 模板


    #include<bits/stdc++.h>
    #define ll long long
    #define re register
    #define gc get_char
    #define cs const
    
    namespace IO{
        inline char get_char(){
            static cs int Rlen=1<<22|1;
            static char buf[Rlen],*p1,*p2;
            return (p1==p2)&&(p2=(p1=buf)+fread(buf,1,Rlen,stdin),p1==p2)?EOF:*p1++;
        }
        
        template<typename T>
        inline T get(){
            char c;
            while(!isdigit(c=gc()));T num=c^48;
            while(isdigit(c=gc()))num=(num+(num<<2)<<1)+(c^48);
            return num;
        }
        inline int getint(){return get<int>();}
    }
    using namespace IO;
    
    using std::cerr;
    using std::cout;
    
    cs int mod=998244353;
    inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
    inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
    inline int mul(int a,int b){static ll r;r=(ll)a*b;return r>=mod?r%mod:r;}
    inline int power(int a,int b,int res=1){
        for(;b;b>>=1,a=mul(a,a))(b&1)&&(res=mul(res,a));
        return res;
    }
    inline void Inc(int &a,int b){(a+=b)>=mod&&(a-=mod);}
    inline void Dec(int &a,int b){(a-=b)<0&&(a+=mod);}
    inline void Mul(int &a,int b){a=mul(a,b);}
    
    typedef std::vector<int> Poly;
    
    std::ostream &operator<<(std::ostream &out,cs Poly &a){
        if(!a.size())out<<"empty ";
        for(int re i:a)out<<i<<" ";
        return out;
    }
    
    cs int bit=20,SIZE=1<<20|1;
    int r[SIZE],*w[bit+1];
    inline void init_NTT(){
        for(int re i=1;i<=bit;++i)w[i]=new int[1<<(i-1)];
        int wn=power(3,mod-1>>bit);
        w[bit][0]=1;for(int re i=1;i<(1<<bit-1);++i)w[bit][i]=mul(w[bit][i-1],wn);
        for(int re i=bit-1;i;--i)
        for(int re j=0;j<(1<<i-1);++j)w[i][j]=w[i+1][j<<1];
    }
    inline void NTT(Poly &A,int len,int typ){
        for(int re i=0;i<len;++i)if(i<r[i])std::swap(A[i],A[r[i]]);
        for(int re i=1,d=1;i<len;i<<=1,++d)
        for(int re j=0;j<len;j+=i<<1)
        for(int re k=0;k<i;++k){
            int &t1=A[j+k],&t2=A[i+j+k],t=mul(t2,w[d][k]);
            t2=dec(t1,t),Inc(t1,t);
        }
        if(typ==-1){
            std::reverse(A.begin()+1,A.begin()+len);
            for(int re i=0,inv=power(len,mod-2);i<len;++i)Mul(A[i],inv);
        }
    }
    inline void init_rev(int l){
        for(int re i=0;i<l;++i)r[i]=r[i>>1]>>1|((i&1)?l>>1:0);
    }
    
    inline Poly operator+(cs Poly &a,cs Poly &b){
        Poly c=a;if(b.size()>a.size())c.resize(b.size());
        for(int re i=0;i<b.size();++i)Inc(c[i],b[i]);
        return c;
    }
    
    inline Poly operator-(cs Poly &a,cs Poly &b){
        Poly c=a;if(b.size()>a.size())c.resize(b.size());
        for(int re i=0;i<b.size();++i)Dec(c[i],b[i]);
        return c;
    }
    
    inline Poly operator*(Poly a,Poly b){
        if(!a.size()||!b.size())return Poly(0,0);
        int deg=a.size()+b.size()-1,l=1;
        if(deg<128){
            Poly c(deg,0);
            for(int re i=0,li=a.size();i<li;++i)
            for(int re j=0,lj=b.size();j<lj;++j)Inc(c[i+j],mul(a[i],b[j]));
            return c;
        }
        while(l<deg)l<<=1;
        init_rev(l);
        a.resize(l),NTT(a,l,1);
        b.resize(l),NTT(b,l,1);
        for(int re i=0;i<l;++i)a[i]=mul(a[i],b[i]);
        NTT(a,l,-1);a.resize(deg);
        return a;
    }
    
    inline Poly Inv(cs Poly &a,int lim){
        Poly c,b(1,power(a[0],mod-2));
        for(int re l=4;(l>>2)<lim;l<<=1){
            c=a;c.resize(l>>1);
            init_rev(l);
            c.resize(l),NTT(c,l,1);
            b.resize(l),NTT(b,l,1);
            for(int re i=0;i<l;++i)b[i]=mul(b[i],dec(2,mul(b[i],c[i])));
            NTT(b,l,-1);b.resize(l>>1);
        }
        b.resize(lim);
        return b;
    }
    
    inline Poly operator/(Poly a,Poly b){
        if(a.size()<b.size())return Poly(0,0);
        int l=1,deg=a.size()-b.size()+1;
        reverse(a.begin(),a.end());
        reverse(b.begin(),b.end());
        while(l<deg)l<<=1;
        b=Inv(b,l);b.resize(deg); 
        a=a*b;a.resize(deg);
        reverse(a.begin(),a.end());
        return a;
    }
    
    inline Poly operator%(cs Poly &a,cs Poly &b){
        if(a.size()<b.size())return a;
        Poly c=a-(a/b)*b;
        c.resize(b.size()-1);
        return c;
    }
    
    inline Poly Ksm(Poly a,int b,Poly mod){
        Poly res(1,1);
        while(b){
            if(b&1)res=res*a%mod;
            a=a*a%mod;
            b>>=1;
        }
        return res;
    }
    namespace BM{
        cs int M=1e4+4;
        int L,cnt,a[M],fail[M],delta[M];
        Poly R[M];
        inline Poly solve(){
            for(int re i=1;i<=L;++i){
                int d=a[i];
                for(int re j=1,lj=R[cnt].size();j<lj;++j)Dec(d,mul(R[cnt][j],a[i-j]));
                if(!d)continue;
                fail[cnt]=i,delta[cnt]=d;
                if(!cnt){++cnt;R[cnt].resize(i+1);}
                else {
                    int coef=mul(delta[cnt],power(delta[cnt-1],mod-2));
                    R[cnt+1].resize(i-fail[cnt-1]);R[cnt+1].push_back(coef);
                    for(int re j=1,lj=R[cnt-1].size();j<lj;++j)R[cnt+1].push_back(mul(mod-coef,R[cnt-1][j]));
                    R[cnt+1]=R[cnt+1]+R[cnt];++cnt;
                }
            }
            return R[cnt];
        }
    }
    
    int n,m;
    Poly f,g;
    signed main(){
    #ifdef zxyoi
        freopen("BM.in","r",stdin);
    #endif
        init_NTT();
        n=getint(),m=getint();BM::L=n;
        for(int re i=1;i<=n;++i)BM::a[i]=getint();
        f=BM::solve();
        for(int re i=1,li=f.size();i<li;++i)cout<<f[i]<<" ";cout<<"
    ";
        std::reverse(f.begin(),f.end());
        for(int re i=0,li=f.size();i<li;++i)f[i]=dec(0,f[i]);f.back()=1;
        g.resize(2);g[1]=1;
        g=Ksm(g,m,f);
        int ans=0;
        for(int re i=0;i<g.size();++i)Inc(ans,mul(g[i],BM::a[i+1]));
        cout<<ans<<"
    ";
        return 0;
    }
    
  • 相关阅读:
    网络拓扑
    OSI 7层模型和TCP/IP 4层模型
    第一范式 第二范式 第三范式 BC范式 第四范式
    医院 信息科
    李纳斯•托瓦兹
    所谓绅士,就是做自己该做之事,而不是想做之事。
    活着
    开头词
    人际题目
    人际关系
  • 原文地址:https://www.cnblogs.com/Inko/p/11747905.html
Copyright © 2020-2023  润新知